Complex Fourier Series: Uncovering the Mystery of Different Results for x^2

In summary, the first post explains the working for the first problem, which is to find the complex Fourier series representation of f(x) = x^2, \ -\pi < x < \pi, \ f(x+2\pi) = f(x) However, the first method returns an undefined answer. The correct calculation involves splitting the complex exponential up into cos and i sin, and proceeding in a very similar fashion, using integration by parts twice.
  • #1
TehAdzMan
6
0
Hello,

First post. I will attempt to use latex, something that involves me jabbing my keyboard with a pen since my \ key is missing.

We have an assignment question which I have solved, but there is a deeper issue I don't understand.

We are asked to find the complex Fourier series representation of

[tex] f(x) = x^2, \ -\pi < x < \pi, \ f(x+2\pi) = f(x) [/tex]

Initially I did the working, used integration by parts twice, and got the result

[tex] f(x) \sim \sum \sinh (in\pi)(\frac{\pi^2 n^2 + 2\pi in+2}{\pi i n^3})e^{-inx}) [/tex]
summed for n between negative infinity to infinity.

This is obviously incorrect because [itex] \sinh (in\pi) = 0 [/itex].

I received a tip off about how to go about solving the question, and early in the working I split the complex exponential term into its cos and isin terms.
Because the region of integration is symmetric and [itex] x^2 \times i\sin(nx) [/itex]
is odd, this term becomes 0 and the complex coefficients become real, the working proceeds and I get the correct answer which is
[tex]f(x) \sim \frac{\pi^2}{3} + 2\sum \frac{(-1)^n}{n^2} e^{-inx} [/tex]

where the sum is over all infinity again, except n = 0 which was worked out seperately.

I've omitted the working because it took so long just to do that, but if I need to I can show it.

I'm pretty sure the working in both cases is right.
What I don't understand is how they come to different answers, excluding for n = 0, which we work out separately, and is the same in either case.
The first way comes up with a sum of 0s, unless I've done something wrong.

I could understand if the first method returned an undefined answer.
Then it is just undefined and we have to try something else to get a defined answer.
But in both cases (excluding n = 0 obviously), the sums are well defined as far as I can tell.

Why are they different? Have I just done something wrong in the working or is there something I don't understand happening?

Adam.
 
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  • #2
Hey TehAdzMan.

Can you show us the working for the first problem?
 
  • #3
Thanks for replying.

Ok so here is the working for the first, incorrect part.

[tex] f(x) \sim \sum^{\infty}_{n = -\infty} C_n e^{-inx}, \ where \ C_n = \frac{1}{2 \pi} \int^{\pi}_{-\pi} x^2 e^{inx} dx \\ [/tex]

Proceeding to calculate Cn using integration by parts

[tex]

C_n = \frac{1}{2 \pi} \{ \left[ \frac {x^2 e^{inx}}{in} \right]^{\pi}_{-\pi} - \int^{\pi}_{-\pi} \frac {2xe^{inx}}{in} dx \} \\

[/tex]

and a second application of integration by parts

[tex]

= \frac{1}{2 \pi} \{ \frac{\pi^2 e^{in \pi} - \pi^2 e^{-in \pi}}{in} - \frac{2}{in} \left[ \left[ \frac {xe^{inx}}{in} \right]^{\pi}_{-\pi} - \int^{\pi}_{ -\pi} \frac{e^{inx}}{in} dx \right] \} \\

= \frac{1}{2 \pi} \{ \frac{\pi^2}{in} (e^{in \pi} - e^{-in \pi}) - \frac{2}{in} \left[ \frac{\pi e^{in \pi}}{in} + \frac{\pi e^{-in \pi}}{in} - \left[ \frac {e^{inx}}{i^2 n^2} \right]^{\pi}_{-\pi} \right] \} \\

= \frac{1}{2 \pi} \{ \frac{2 \pi^2}{in} \sinh (in \pi) - \frac{4 \pi}{i^2 n^2} \sinh (in \pi) - \frac{2}{i^3 n^3} (e^{in \pi} - e^{-in \pi}) \} [/tex]

which, with some arrangement gives

[tex] C_n = \sinh (in\pi)(\frac{\pi^2 n^2 + 2\pi in+2}{\pi i n^3}) [/tex]

So basically sinh(in pi) is coming out at every step, which = 0.

The correct calculation just involves splitting the complex exponential up into cos and i sin, getting rid of the i sin for the aforementioned reason, and proceeding in a very similar fashion, using integration by parts twice.

Is something wrong in that working?
 
  • #4
Ok there was a working error. My bad.
It turns out it is the same.

I learned latex tho so that's good.
I'm trying to work out how to delete this or mark it as DONT READ or something now.
 
  • #5
Why can I only edit my last post? Can I change the thread name etc?
Can I delete the post to rid this forum of a useless post?
I looked for this info but couldn't find it.
 

Related to Complex Fourier Series: Uncovering the Mystery of Different Results for x^2

1. What is a Complex Fourier Series?

A Complex Fourier Series is a mathematical representation of a periodic function as a sum of complex exponential functions. It is used to decompose a complex function into simpler components, making it easier to analyze and understand.

2. How is a Complex Fourier Series different from a regular Fourier Series?

A Complex Fourier Series uses complex exponential functions, while a regular Fourier Series uses only real sine and cosine functions. This allows for a more general representation of periodic functions.

3. What is the main application of Complex Fourier Series?

Complex Fourier Series are commonly used in signal processing and electrical engineering to analyze and manipulate periodic signals. They are also used in other fields such as physics, mathematics, and economics.

4. Can any function be represented by a Complex Fourier Series?

No, not every function can be represented by a Complex Fourier Series. The function must be periodic and have a finite number of discontinuities for it to be represented accurately.

5. What are the limitations of using a Complex Fourier Series?

While Complex Fourier Series can accurately represent many periodic functions, they may not be able to accurately represent functions with sharp, localized changes. In these cases, other mathematical tools and techniques may be more appropriate.

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